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Question Number 192756 by mokys last updated on 26/May/23

Commented by mokys last updated on 26/May/23

$$Imcantsolvethisbecauseitsveryhardcanhelpsmepleas?$$

Answered by witcher3 last updated on 02/Jun/23

$${\mathrm{y}}^{\prime }={\mathrm{y}}^{\left(1\right)}\mathrm{for}\mathrm{all}\mathrm{the}\mathrm{rest}$$$${\mathrm{y}}^{\left(3\right)}-{2\mathrm{x}}^2{\mathrm{y}}^{″}+{5\mathrm{xy}}^{\prime }+3\mathrm{y}=\cos \left(\mathrm{x}\right)$$$$\mathrm{Suppose}\mathrm{\exists }({\mathrm{y}}_1,{\mathrm{y}}_2)\in {\mathrm{C}}_3^2\mathrm{such}\mathrm{that}{\mathrm{y}}_1\mathrm{and}{\mathrm{y}}_2\mathrm{are}$$$$\mathrm{solution}$$$$\Rightarrow \mathrm{f}={\mathrm{y}}_1-{\mathrm{y}}_2\mathrm{is}\mathrm{solution}\mathrm{of}$$$${\mathrm{f}}^{\left(3\right)}-{2\mathrm{x}}^2{\mathrm{f}}^{\left(2\right)}+{5\mathrm{xf}}^{\left(1\right)}+3\mathrm{f}=0....\left(\mathbf{E}\right)$$$$\mathrm{f}(-2)={\mathrm{f}}^{\prime }(-2)={\mathrm{f}}^{\left(2\right)}(-2)=0$$$$\mathrm{applied}\left(\mathrm{E}\right)\Rightarrow {\mathrm{f}}^{\left(3\right)}=0$$$$\mathrm{by}\mathrm{reccursion}$$$${\mathrm{f}}^{\left(3\right)}={2\mathrm{x}}^2{\mathrm{f}}^{{}^{″}}-{5\mathrm{xf}}^{\prime }-3\mathrm{f}\Rightarrow \mathrm{f}\in {\mathrm{C}}_4$$$$$$$$\mathrm{reccurssiv}\Rightarrow \mathrm{f}\in {\mathrm{C}}_{\mathrm{\infty }}$$$$\mathrm{and}\mathrm{\forall }\mathrm{n}\in \mathbb{N}{\mathrm{f}}^{\left(2\right)}(-2)=0$$$${\mathrm{f}}^{\left(\mathrm{n}\right)}\left(\mathrm{x}\right)=\underset{\mathrm{i}=0}{\sum ^{\mathrm{n}}}\mathrm{p}\left(\mathrm{x}\right){\mathrm{f}}^{\left(\mathrm{i}\right)}\left(\mathrm{x}\right),\mathrm{\exists }\mathrm{p}\in {\mathbb{R}}^2\left[\mathrm{X}\right]$$$$\mathrm{applie}\mathrm{taylor}\mathrm{arround}(-2)$$$$\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{f}}^{\left(\mathrm{k}\right)}(-2)}{\mathrm{k}!}{(\mathrm{x}+2)}^{\mathrm{k}}$$$$\mathrm{but}{\mathrm{f}}^{\left(\mathrm{k}\right)}\left(2\right)=0,\mathrm{\forall }\mathrm{k}\in \mathbb{N}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=0$$$${\mathrm{y}}_1-{\mathrm{y}}_2=0\Rightarrow {\mathrm{y}}_1={\mathrm{y}}_2$$$$\mathrm{\exists }\mathrm{unique}\mathrm{solution}$$$$$$$$$$

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